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An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude $2.0 × 10^4 N C^{–1} [Fig.(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. (b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

1 Answer

In the figure (a), the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field.
Acceleration of the electron, ae = eE/me, where me is the mass of the electron
Starting from rest, the time required by the electron to fall through a distance h is given by $t_e= \sqrt {\large\frac{2h}{a_e}}$
$\qquad= \sqrt {\large\frac{2hm}{2E}}$
For $e = 1.6 × 10^{–19} \;C, m_e = 9.11 × 10^{–31}\; kg$
$E= 2.0 \times 10^4 NC^{-1}, h= 1.5 \times 10^{-2}m, t_e =2.9 \times 10^{-9}$
In the figure (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is
$a_p= eE/m_p$
where $m_p$ is the mass of the proton; $mp = 1.67 × 10–27 kg. The time of fall for the proton is
$t_p =\sqrt {\large\frac{2h}{a_p} }$
$\quad= \sqrt {\large\frac{2hm_p}{eE}}$
$\quad = 1.3 \times 10^{-7}\;s$
Thus, the heavier particle (proton) takes a greater time to fall through the same distance.
This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body.
Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall.
To see if this is justified, let us calculate the acceleration of the proton in the given electric field:
$a_p =\large\frac{(1.6 \times 10^{-19} C ) \times (2.0 \times 10^4 NC^{-1})}{1.67 \times 10^{-27}\;kg}$$=1.9 \times 10^{12}\;ms^{-2}$
which is enormous compared to the value of $g(9.8 ms^{-2} )$,the acceleration due to gravity.
The acceleration of the electron is even greater.
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